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a: f(x)=3x^4+2x^3+6x^2-x+2

g(x)=-3x^4-2x^3-5x^2+x-6

b: H(x)=f(x)+g(x)

=3x^4+2x^3+6x^2-x+2-3x^4-2x^3-5x^2+x-6

=x^2-4

f(x)-g(x)

=3x^4+2x^3+6x^2-x+2+3x^4+2x^3+5x^2-x+6

=6x^4+4x^3+11x^2-2x+8

c: H(x)=0

=>x^2-4=0

=>x=2 hoặc x=-2

a) \(P\left(x\right)=3x^3-x^2-2x^4+3+2x^3+x+3x^4-x^2-2x^4+3+2x^3+x+3x^4\)

 \(=2x^4+7x^3-2x^2+2x+6\)

\(Q\left(x\right)=-x^4+x^2-4x^3-2+2x^2-x-x^3-x^4+x^2-4x^3-2+2x^2-x-x^3\)

\(=-2x^4-10x^3+6x^2-2x-4\)

b) \(P\left(x\right)+Q\left(x\right)=2x^4+7x^3-2x^2+2x+6-2x^4-10x^3+6x^2-2x-4\)

                                      \(=-3x^3+4x^2+2\)

`@` `\text {dnv4510}`

`A)`

`P(x)+Q(x)=`\((2x^4+3x^2-3x^2+6)+(x^4+x^3-x^2+2x+1)\)

`= 2x^4+3x^2-3x^2+6+x^4+x^3-x^2+2x+1`

`= (2x^4+x^4)+x^3+(3x^2-3x^2-x^2)+2x+(6+1)`

`= 3x^4+x^3-x^2+2x+7`

`B)`

`P(x)+M(x)=2Q(x)`

`-> M(x)= 2Q(x) - P(x)`

`2Q(x)=2(x^4+x^3-x^2+2x+1)`

`= 2x^4+2x^3-2x^2+4x+2`

`-> 2Q(x)-P(x)=(2x^4+2x^3-2x^2+4x+2)-(2x^4+3x^2-3x^2+6)`

`= 2x^4+2x^3-2x^2+4x+2-2x^4-3x^2+3x^2-6`

`= (2x^4-2x^4)+2x^3+(-2x^2-3x^2+3x^2)+4x+(2-6)`

`= 2x^3-2x^2+4x-4`

Vậy, `M(x)=2x^3-2x^2+4x-4`

`C)`

Thay `x=-4`

`M(-4)=2*(-4)^3-2*(-4)^2+4*(-4)-4`

`= 2*(-64)-2*16-16-4`

`= -128-32-16-4`

`= -180`

`->` `x=-4` không phải là nghiệm của đa thức.

1: \(A=5x^5-5x^3+7x^2-2x+4\)

\(B\left(x\right)=-5x^6+2x^4+4x^3+4x^2-4x-1\)

2: \(A\left(x\right)+B\left(x\right)=5x^5-5x^3+7x^2-2x+4-5x^6+2x^4+4x^3+4x^2-4x-1\)

\(=-5x^6+5x^5+2x^4-x^3+11x^2-6x+3\)

\(A\left(x\right)-B\left(x\right)\)

\(=5x^5-5x^3+7x^2-2x+4+5x^6-2x^4-4x^3-4x^2+4x+1\)

\(=5x^6+5x^5-2x^4-9x^3+3x^2+2x+5\)

\(F\left(x\right)=3x^4+2x^3+6x^2-x+2\)

\(G\left(x\right)=-3x^4-2x^3-5x^2+x-6\)

F(x)=-x+2+5x²+2x⁴+2x³+x²+x⁴

F(x)= ( 5x²+x²) + ( 2x⁴ +x⁴)  +2x³-x+2

F (x) = 6x² + 3x⁴ +2x³-x+2

G(x) = -x²+x³+x-6-3x³-4x²-3x⁴

G (x) = ( -x² -4x²) + ( x³ -3x³) -3x⁴ +x-6

G (x) =  -5x² - 2x³ -3x⁴ +x-6

Bài 1:

\(a,x^4+5x^2+9\\=(x^4+6x^2+9)-x^2\\=[(x^2)^2+2\cdot x^2\cdot3+3^2]-x^2\\=(x^2+3)^2-x^2\\=(x^2+3-x)(x^2+3+x)\)

\(b,x^4+3x^2+4\\=(x^4+4x^2+4)-x^2\\=[(x^2)^2+2\cdot x^2\cdot2+2^2]-x^2\\=(x^2+2)^2-x^2\\=(x^2+2-x)(x^2+2+x)\)

\(c,2x^4-x^2-1\\=2x^4-2x^2+x^2-1\\=2x^2(x^2-1)+(x^2-1)\\=(x^2-1)(2x^2+1)\\=(x-1)(x+1)(2x^2+1)\)

Bài 2:

\(a,\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=120\)

\(\Leftrightarrow\left[\left(x+1\right)\left(x+4\right)\right]\cdot\left[\left(x+2\right)\left(x+3\right)\right]=120\)

\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=120\) (1)

Đặt \(x^2+5x+5=y\), khi đó (1) trở thành:

\(\left(y-1\right)\left(y+1\right)=120\)

\(\Leftrightarrow y^2-1=120\)

\(\Leftrightarrow y^2=121\)

\(\Leftrightarrow\left[{}\begin{matrix}y=11\\y=-11\end{matrix}\right.\)

+, TH1: \(y=11\Leftrightarrow x^2+5x+5=11\)

\(\Leftrightarrow x^2+5x-6=0\)

\(\Leftrightarrow x^2-x+6x-6=0\)

\(\Leftrightarrow x\left(x-1\right)+6\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-6\end{matrix}\right.\left(\text{nhận}\right)\)

+, TH2: \(y=-11\Leftrightarrow x^2+5x+5=-11\)

\(\Leftrightarrow x^2+5x+16=0\)

\(\Leftrightarrow\left[x^2+2\cdot x\cdot\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2\right]-\dfrac{25}{4}+16=0\)

\(\Leftrightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}=0\)

Ta thấy: \(\left(x+\dfrac{5}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}\ge\dfrac{39}{4}>0\forall x\)

Mà \(\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}=0\)

\(\Rightarrow\) loại

Vậy \(x\in\left\{1;-6\right\}\).

\(b,\) Đề thiếu vế phải rồi bạn.

Thu gọn Q(x) = x4 + 7x2 + 1

Khi đó R(x) = Q(x) - P(x) = 4x2 + 3x + 2. Chọn A

thu gọn rồi chứng minh nó > 0

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